In Tom Apostol Calculus: method of exhaustion for the area of a parabolic segment of $x^2$, proving $b^3/3$ is the only number between $s_n$ and $S_n$

In the method of exhaustion for the area of a parabolic segment of $$x^2$$, the part where the author proves that $$b^3/3$$ is the only number which satisfies $$s_n

The prove has come down to this

$$\frac{b^3}{3}-\frac{b^3}{n}< A<\frac{b^3}{3}+\frac{b^3}{n}$$ for every integer $$n>=1$$. ,.....(1)

Now there are three possibilities

$$A>\frac{b^3}{3}$$ , $$A<\frac{b^3}{3}$$ , $$A<\frac{b^3}{3}$$

The author now shows that the first two inequalities lead to a contradiction, then we must have $$A=\frac{b^3}{3}$$.

(what I find difficult to understand is the contradiction, I can't see which fact is being contradicted.)

The author says this:

Suppose the inequality $$A>\frac{b^3}{3}$$ were true. Then from the second inequality in (1) we obtain

$$A-\frac{b^3}{3}<\frac{b^3}{n}$$ for every integer $$n \geq 1$$.

Since $$A-\frac{b^3}{3}$$ is positive, we may divide both side by $$A-\frac{b^3}{3}$$ and then multiply by n to obtain the equivalent statement $$n<\frac{b^3}{A-\frac{b^3}{3}}$$ for every n.

But this inequality is obviously false when $$n \geq \frac{b^3}{A-\frac{b^3}{3}}$$. Hence the inequality $$A>\frac{b^3}{3}$$ leads to contradiction.

Is it saying that $$A>\frac{b^3}{3}$$ is true only when $$n<\frac{b^3}{A-\frac{b^3}{3}}$$. and it contradicts from the assumption that it should be true for all $$n$$?
You have showed that, if we assume that $$A > \frac{b^3}{3}$$, then equation (1) is true only when $$n < \frac{b^2}{A - \frac{b^3}{3}}$$. However, the equation (1) must be true for all integers $$n \geq 1$$. We have reached a contradiction since our assumption implies an upper bound for $$n$$ in equation (1), but the equation must be true for all integer $$n \geq 1$$.